Problem: A secant line intersects the graph of $y=\cos(x)$ at two points with $x$ -coordinates $6$ and $t$. What is the slope of the secant line? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\cos(t-6)}{t-6}$ (Choice B) B $\dfrac{\cos(t-6)}{t}$ (Choice C) C $\dfrac{\cos(t)-\cos(6)}{t-6}$ (Choice D) D $\dfrac{\cos(t)-\cos(6)}{6}$
We are given that the secant line intersects the graph at $x=6$ and $x=t$. Since these points are on the curve $y=\cos(x)$, we know that their $y$ -values are $y=\cos(6)$ and $y=\cos(t)$, respectively. To summarize this part, we know that the secant line passes through the points $(6,\cos(6))$ and $(t,\cos(t))$. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{\cos(t)-\cos(6)}{t-6} \end{aligned}$ In conclusion, the slope of the secant line is $\dfrac{\cos(t)-\cos(6)}{t-6}$.